Question

Help please on this question

Answer 1

Given,

`\sin (7x+1)=\cos (x+33)`

We know that,

`\sin \theta=\cos (90-\theta)`
`\begin{gathered} \cos (90-(7x+1))=\cos (x+33) \\ \text{equating the parameters of cosines on both sides we get,} \\ 90-7x-1=x+33 \\ 89=7x+x+33 \\ 8x=89-33 \\ 8x=56 \\ x=(56)/(8) \\ x=7 \end{gathered}`
`\begin{gathered} \text{Angle 1 =(7x+1)} \\ \text{Angle 1 =(7}*7)\text{+1)} \\ \text{Angle 1 =}49+1 \\ \text{Angle 1 =5}0^(\circ) \end{gathered}`
`\begin{gathered} \text{Angle 2 =(}x+33\text{)} \\ \text{Angle 2 =(}7+33\text{)} \\ \text{Angle 2 = 40}^(\circ) \end{gathered}`